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Question

Answers

A. Resistance

B. Inductance

C. Capacitance

D. Magnetic flux

Answer
Verified

Hint: To solve this question first we will find the SI unit of the given quantities in the option. We can do this simply by writing the mathematical expression or by breaking them into fundamental quantities. Then compare them with the SI unit given in the question to find the required answer.

__Complete step by step answer:__

SI unit or International System of Unit is based on meter, kilogram, second, ampere, kelvin candela and mole. It uses a set of prefixes to indicate the multiplication and division in terms of power of 10.

Resistance can be defined from Ohm’s law as,

$R=\dfrac{V}{I}$

Voltage can be defined as the product of electric field and distance. Electric field is defined in terms of force per unit charge. Dimension of charge is given by product of current and time.

So,

$R=\dfrac{V}{I}=\dfrac{\dfrac{Fd}{q}}{I}=\dfrac{Fd}{qI}=\dfrac{Fd}{It\times I}=\dfrac{Fd}{{{I}^{2}}t}$

Now, SI unit of force is, $kgm{{s}^{-2}}$

Si unit of current is A, SI unit of time is second and SI unit of distance is meter.

So, from the above equation we can write that the SI unit of resistance is,

$\dfrac{kgm{{s}^{-2}}m}{{{A}^{2}}s}=kg{{m}^{2}}{{A}^{-2}}{{s}^{-3}}$

Now, inductance is given by the mathematical expression as,

From above, the SI unit of voltage can be given as, $kg{{m}^{2}}{{A}^{-1}}{{s}^{-3}}$

SI unit of current is A and time is second.

Si unit of inductance will be, $\dfrac{kg{{m}^{2}}{{A}^{-1}}{{s}^{-3}}}{\dfrac{A}{s}}=kg{{m}^{2}}{{A}^{-2}}{{s}^{-2}}$

Capacitance can be defined as,

$C=\dfrac{Q}{V}=\dfrac{It}{V}$

So, the SI unit of capacitance will be, $\dfrac{As}{kg{{m}^{2}}{{A}^{-1}}{{s}^{-3}}}=k{{g}^{-1}}{{m}^{-2}}{{A}^{2}}{{s}^{4}}$

Magnetic flux can be given by the product of magnetic field and area.

$\phi =BA$

Magnetic field is given by Lorentz force as,

$B=\dfrac{F}{qv}=\dfrac{F}{Itv}$

The SI unit of magnetic field will be,

$\dfrac{kgm{{s}^{-2}}}{Asm{{s}^{-1}}}=kg{{A}^{-1}}{{s}^{-2}}$

So, SI unit of magnetic flux will be $kg{{A}^{-1}}{{s}^{-2}}\times {{m}^{2}}=kg{{m}^{2}}{{A}^{-1}}{{s}^{-2}}$

From the above calculation we find that SI unit of resistance is $kg{{m}^{2}}{{A}^{-2}}{{s}^{-3}}$

The correct option is (A)

Note: Sometimes we can express the derived physical quantities with more than one formula where we have different physical quantities. We can use any one of them. For example, if we consider the inductance, we can express it as magnetic flux per unit current flowing. If we take this expression also, we will get the same answer.

SI unit or International System of Unit is based on meter, kilogram, second, ampere, kelvin candela and mole. It uses a set of prefixes to indicate the multiplication and division in terms of power of 10.

Resistance can be defined from Ohm’s law as,

$R=\dfrac{V}{I}$

Voltage can be defined as the product of electric field and distance. Electric field is defined in terms of force per unit charge. Dimension of charge is given by product of current and time.

So,

$R=\dfrac{V}{I}=\dfrac{\dfrac{Fd}{q}}{I}=\dfrac{Fd}{qI}=\dfrac{Fd}{It\times I}=\dfrac{Fd}{{{I}^{2}}t}$

Now, SI unit of force is, $kgm{{s}^{-2}}$

Si unit of current is A, SI unit of time is second and SI unit of distance is meter.

So, from the above equation we can write that the SI unit of resistance is,

$\dfrac{kgm{{s}^{-2}}m}{{{A}^{2}}s}=kg{{m}^{2}}{{A}^{-2}}{{s}^{-3}}$

Now, inductance is given by the mathematical expression as,

From above, the SI unit of voltage can be given as, $kg{{m}^{2}}{{A}^{-1}}{{s}^{-3}}$

SI unit of current is A and time is second.

Si unit of inductance will be, $\dfrac{kg{{m}^{2}}{{A}^{-1}}{{s}^{-3}}}{\dfrac{A}{s}}=kg{{m}^{2}}{{A}^{-2}}{{s}^{-2}}$

Capacitance can be defined as,

$C=\dfrac{Q}{V}=\dfrac{It}{V}$

So, the SI unit of capacitance will be, $\dfrac{As}{kg{{m}^{2}}{{A}^{-1}}{{s}^{-3}}}=k{{g}^{-1}}{{m}^{-2}}{{A}^{2}}{{s}^{4}}$

Magnetic flux can be given by the product of magnetic field and area.

$\phi =BA$

Magnetic field is given by Lorentz force as,

$B=\dfrac{F}{qv}=\dfrac{F}{Itv}$

The SI unit of magnetic field will be,

$\dfrac{kgm{{s}^{-2}}}{Asm{{s}^{-1}}}=kg{{A}^{-1}}{{s}^{-2}}$

So, SI unit of magnetic flux will be $kg{{A}^{-1}}{{s}^{-2}}\times {{m}^{2}}=kg{{m}^{2}}{{A}^{-1}}{{s}^{-2}}$

From the above calculation we find that SI unit of resistance is $kg{{m}^{2}}{{A}^{-2}}{{s}^{-3}}$

The correct option is (A)

Note: Sometimes we can express the derived physical quantities with more than one formula where we have different physical quantities. We can use any one of them. For example, if we consider the inductance, we can express it as magnetic flux per unit current flowing. If we take this expression also, we will get the same answer.

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